91Ó°ÊÓ

Given data:

The following are descriptive statistics and a histogram:

Shark lengths are nearly symmetric, peaking at $16$, and range from $9.40$feet to $22.8$feet.

Shape: symmetric and roughly bell-shaped.

Center: Mean is $15.5884$ and median is $15.75$.

Spread: standard deviation is $2.5499$.

Given data:

Mean, $\stackrel{¯}{x}=15.586$

Sample standard deviation, $s=2.550$

$\stackrel{¯}{x}-s=13.036$

$\stackrel{¯}{x}+s=18.136$

Percent of interval

$(68.26\%)68.2\%$

$\stackrel{¯}{x}-2s=10.487$

$\stackrel{¯}{x}+2s=20.686$

Percent of interval

$(95.44\%)95.5\%$

$\stackrel{¯}{x}-3s=7.937$

$\stackrel{¯}{x}+3s=23.236$

Percent of interval

$(99.73\%)100.0\%$

According to the $68-95-99.7$rule, approximately $68.2\%$of the data values fall within one standard deviation. $95.5\%$of the data values fall within two standard deviations, and $100\%$ of the data values fall within three standard deviations of the mean.

Given data:

The following is a Minitab normal probability plot:

Except for one small and one huge shark length, the figure is reasonably linear, demonstrating that the Normal distribution is reasonable.

Given data:

The results of parts (a), (b), and (c) show that shark lengths are about average.

The histogram was roughly bell-shaped, and the normal probability plot was roughly normal.